= f x 1 + h. 3. Geometrically, the average rate of change is the slope of the secant line connecting the pts (x 1 )).


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1 Math 1205 Calculus/Sec. 3.3 The Derivative as a Rates of Change I. Review A. Average Rate of Change 1. The average rate of change of y=f(x) wrt x over the interval [x 1, x 2 ]is!y!x ( )  f( x 1 ) = y 2 " y 1 = f x 2 x 2 " x 1 x 2  x 1 2. Difference Quotient: f x 1 + h ( )  f ( x 1 ) h ( )  f( x 1 ) = f x 1 + h h,where h = x 2  x 1,where h = x 2  x Geometrically, the average rate of change is the slope of the secant line connecting the pts (x 1, f (x 1 )) and (x 2, f (x 2 )). 4. Physically, the average rate of change is the average velocity. B. (Instantaneous) Rate of Change f (x 1. The rate of change of y=f(x) wrt x at x = x 0 is f!(x 0 ) = lim 0 + h) # f (x 0 ) h"0 h the limit exist. 2. Derivative at a point x = a : f!(a) = lim h"0 f (a + h) # f (a) h, provided the limit exists. 3. Geometrically, the rate of change is the slope of the tangent line at x = a. 4. Physically, the rate of change is the velocity at x = a., provided II. Physics A. Rectilinear motion / Free Fall (Air resistance will be ignored) 1. Position Function: s(t) position as a function of time of a particle that is moving in a straight line. 2. Displacement:! s = s( t +!t) " s( t) (change in position Not total distance) 3. Velocity a. Average velocity of an object over a time interval is ( ) " s( t)!s!t = s t +!t!t ( ) " s( t 1 ) = s t 2 t 2 " t 1 b. Velocity (Instantaneous velocity) is the rate of change of displacement,!s with respect to time. Velocity is the derivative of position with respect to time. c. Formula: If a body s position at time t is s t v( t) = ds dt = s! t ( ) = lim "t#0 $ "s ' s t + "t % & "t ( ) = lim "t#0 "t ( ), then the body s velocity is ( ) * s( t)
2 d. Since the derivative graphically represents the slope of the tangent line at t 0, 1.) m tan > 0! s "(t 0 ) > 0! v(t 0 ) > 0! graph of s(t) is rising! position is increasing! object is moving to the right / up. 2.) m tan < 0! s "(t 0 ) < 0! v(t 0 ) < 0! graph of s(t) is falling! position is decreasing! object is moving to the left / down. 3. Speed is the absolute value of velocity. v( t) = ds dt ; Speed >0!! 4. Acceleration a. Acceleration is the derivative of velocity with respect to time. b. Formula: If a body s position at time t is s( t), then the body s acceleration at time t is a( t) = dv dt = d 2 s dt = s!! t 2 ( ) = lim "t#0 $ % & "v "t ' ( ) c. When a(t) > 0! velocity is increasing d. When a(t) < 0! velocity is decreasing " v(t) > 0! object speeds up $ # %$ v(t) < 0! object slows down " v(t) > 0! object slows down $ # %$ v(t) < 0! object speeds up e. Summary of the relationship between the signs of velocity and acceleration When velocity and acceleration have the same algebraic sign, the object speeds up. When velocity and acceleration have the different algebraic signs, the object slows down. 5. Jerk is the derivative of acceleration with respect to time: j(t) = a! ( t) = da dt = s!!! t ( ) = d 3 s dt 3 III. Examples A. Suppose s(t) = t 2! 3t + 4 gives the position of a body moving on a coordinate line for 0! t! 4, where t is measured in seconds and s in meters. 1. Determine the position of the particle at the endpoints. 2. Determine the displacement.
3 3. Determine the average velocity of the particle over the interval [0,6] 4. Determine velocity after 1s and after4s. 5. Determine when the particle changes direction. 6. Determine when the particle is moving forward (i.e., in a positive direction). 7. Determine the total distance traveled by the particle on the interval [0,4]. 8. Draw a diagram to represent the motion of the particle.
4 B. A projectile is fired straight up from the ground. Its distance from the ground after t seconds is: s(t) =!16t t (measured in feet) 1. Determine the velocity and the acceleration functions. 2. Determine the velocity at t= 4 sec and at t=9sec. 3. Determine when the projectile reaches its maximum height. 4. Determine the maximum height the projectile obtains. 5. Determine when the projectile hits the ground. 6. Determine the impact velocity of the projectile.
5 C. Given the position function s t ( ) =! 1 2 t 2 + cos( 3t) for 0! t! 3 meters and t is measured in seconds. 1. Determine the velocity, acceleration and jerk as a function of time., where s is measured in 2. Determine the position, velocity, acceleration and jerk at t =! 4 s 3. Given the graph below, mark up the graph to indicate the following c. Where is the body located on the axis at the beginning of the time interval? at the end? b. When is the body momentarily at rest? c. When does it move to the left? to the right? d. When does it speed up? slow down? e. When is it moving fastest? slowest? f. When is it farthest from the axis origin?
6 D. A rock is tossed straight up from a cliff 112 ft above the ground. Its distance from the ground below after t seconds is s(t) =!16t t (measured in ft) 1. Determine the velocity and the acceleration functions. 2. Determine the initial velocity of the rock 3. Determine the maximum height the rock obtains. 4. Determine the rock s impact velocity. E. The quantity of charge Q in coulombs (C) that has passed through a pt in a wire up to time t (measured in seconds) is given by Q(t) = t 3! 2t 2 + 6t + 2. Find the current I, when dq t = 1 2 s if the current I is the rate at which charge flows through a surface, i.e., I = dt.
7 F. A spherical balloon is being inflated. Find the rate of increase of the surface area wrt the radius r when r is 2 ft. G. Graphical Examples In order to identify the function from the 1 st derivative and the 2 nd derivative, look for the following. Notice that f! = 0 whenever the fn has a horizontal tangent. Also f! > 0 when the fn has a positive slope and! information to distinguish f! from f!!, i.e., f!! = 0 whenever y = f! has a horizontal tangent, etc. f < 0 when the fn has a negative slope. We can also use this Given the graphs below, determine s(t), v(t) and a(t), i.e., f, f! and f!!
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